You have found the following ages (in years) of 4 bears. The bears are randomly selected from the 22 bears at your local zoo: $ 5,\enspace 4,\enspace 6,\enspace 39$ Based on your sample, what is the average age of the bears? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 22 bears, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{72.25} + {90.25} + {56.25} + {650.25}} {{4 - 1}} $ $ {s^2} = \dfrac{{869}}{{3}} = {289.67\text{ years}^2} $ We can estimate that the average bear at the zoo is 13.5 years old. There is a variance of 289.67 years $^2$.